BMW-M-Mexico reports nearly zero difference at the track with 20s. I run 20s currently. The only reason I want a set of 19s for track is for the better tire choices. There aren't a lot of good tires in the 20" size. 18s have the most options but 18s won't fit my BBK.

BMW-M-Mexico reports nearly zero difference at the track with 20s. I run 20s currently. The only reason I want a set of 19s for track is for the better tire choices. There aren't a lot of good tires in the 20" size. 18s have the most options but 18s won't fit my BBK.

No offense to either of you but really, 20s are not for trackwork (unless you have a GTR, and even they're looking for small track wheels). One anecdotal trackday with no data and an unknown driver is not exactly proof.

20" wheels and tires on the Turner car are for show, they want to sell you big wheels with enormous staggers, but I seriously doubt they would sell you this for any sort of performance advantage on the track or even on the street.

Tires look like Pirelli P Zero Neros, not the greatest rubber.

I never said anything about tracking anything. This is a street car.

See post #15 for my thoughts.

And if you think Turner is into just selling big wheels with enormous staggers, then you are wrong.

Can someone please explain to me why this 20" rim setup wouldn't perform as well as the 19" rim setup detailed below? Let me explain that both the 19's & 20's are the same rim and same tire brand just different sizes.

Here is my conclusion. The 20" rim setup will perform better for 2 reasons. 1) The overall wheel & tire weight is 2lbs less. Not a big difference but a difference none the less.
2) Since the 20" tires weigh less than the 19's, even if the overall weight were the same, the 20's would perform better because there is less weight farther from the center of rotation. Since the overall tire diameter is roughly the same even though the rims are heavier the tire is the farthest point from the center. Having a lighter tire will perform better because of less rotational mass. Plus the overall package being lighter.

I'm no engineer but it logically makes sense to me that when you distribute the weight closer to the center of rotation it improves performance. Does this make sense? Thanks for any and all suggestions.

What is more dense? Rubber or metal? With 20s you are putting more metal at a farther distance from the center, hence higher rotational inertia (I) even though the tires weigh a bit less.

With the same design, 20s will for sure have higher "I" than 19s. Period.

Now, how much impact does that extra "I" have on the performance of the car? I don't know. Probably need to do some math to figure out.

Now, how much impact does that extra "I" have on the performance of the car? I don't know. Probably need to do some math to figure out.

I bet it isn't much at all. Hence Nissan put 20's on the GT-R. They must not be big on the whole inertia thing. And if you put 20's on that are 3-4 lbs per/wheel lighter than the stock 19's. It would have to offset some/most of the negative affect.

I bet it isn't much at all. Hence Nissan put 20's on the GT-R. They must not be big on the whole inertia thing. And if you put 20's on that are 3-4 lbs per/wheel lighter than the stock 19's. It would have to offset some/most of the negative affect.

The GT-R is a REALLY big car, 20s on that car are very much like 18s or 19s on an M3. There is a reason you fight I EVERYWHERE on a performance car. If you don't understand the effect of inertia you really need to grab a Physics 101 text book. I assure you Nissan engineers understand the importance of I.

As briefly as possble rotating bodies obey an equation that is a perfect equivalent of Newtons law, force = mass x acceleration. Less mass, same force = larger acceleration. For rotating bodies torque = moment of inertia x angular acceleration. Again low I is absolutely fundamental to making objects easy to "spin up" (i.e. have a high angular acceleration for a give applied torque).

All rotating masses have to be accelerated linearly along with the translational movement of the car, this takes force and power. As well rotating objects have to not only be accelerated they have to be spun up, this takes even more torque and power. Surely you need to fight weight as well, both are key. Large I values in your wheels not only affect acceleration but also braking and turning, all negatively. In other places that contain rotating mechanical parts like an engine, axles, driveshafts, etc. it is always a balance to have low weight, low I and acceptable stiffness and strength. For these components low mass and I also produce much less bearing loads.

There is a similar reason to be greatly concerned with unsprung mass - that of the suspension and everything that moves along with the suspension. High unsprung mass is even worse than high sprung mass (say like the car body or a passenger).

Lastly (and again, and again...) these effects are NOT huge. We are probably talking tenths of seconds in most acceleration metrics and maybe a couple feet in braking metrics will separate 18s, 19s and 20s. Very rough educated guesses here. However, do recall that saving X lb in a rotating component is MUCH more significant than saving the same X lb from a non-rotating component.

BMW-M-Mexico reports nearly zero difference at the track with 20s. I run 20s currently. The only reason I want a set of 19s for track is for the better tire choices. There aren't a lot of good tires in the 20" size. 18s have the most options but 18s won't fit my BBK.

I don't understand how the 20's can be fine considering you run a 305 rear and a 255 front which is way off from the balance of the stock 245/265 . Is the car more prone to sudden oversteer with the 20's due to the front / rear size difference.?

Large I values in your wheels not only affect acceleration but also braking and turning, all negatively.

But of course this doesn't apply to the 20's on the GT-R. Because it is big, special, and was designed for 20's. The magic electronics and suspension on the car just eliminate inertia from the equation. I am surprised Nissan hasn't figured out how to eliminate the force a gravity while they are at it. Fact is Nissan picked 20's for the GT-R.

Quote:

Originally Posted by swamp2

There is a similar reason to be greatly concerned with unsprung mass - that of the suspension and everything that moves along with the suspension. High unsprung mass is even worse than high sprung mass (say like the car body or a passenger).

Glad to see you agree. 20's that weigh 3-4 lbs lighter per/wheel than the stock 19's, will have less unsprung weight/mass.

Quote:

Originally Posted by swamp2

Lastly (and again, and again...) these effects are NOT huge. We are probably talking tenths of seconds in most acceleration metrics and maybe a couple feet in braking metrics will separate 18s, 19s and 20s. Very rough educated guesses here. However, do recall that saving X lb in a rotating component is MUCH more significant than saving the same X lb from a non-rotating component.

Again you agree, these effects are not huge.

I will say it again. 20's for the street and a seperate 18 inch track setup. It isn't that difficult.

But of course this doesn't apply to the 20's on the GT-R. Because it is big, special, and was designed for 20's. The magic electronics and suspension on the car just eliminate inertia from the equation. I am surprised Nissan hasn't figured out how to eliminate the force a gravity while they are at it. Fact is Nissan picked 20's for the GT-R.

Sarcasm will not make up for you seeming lack of understanding. Of course it applies. I is I is I, period. The OEM choice of a tire wheel size combination is one subject to a great many variables and compromises. If you could replace the wheels on the GT-R with wheels that have a smaller I it would perform better, period, no if's and's or but's. BUT given the gearing, brake sizes, existing suspension design, etc. you (almost for sure) can't find 19s that will work on the car. If you could you'd likely have tires with sidewalls too flexible for the suspension. I suspect brake rotor diameter is actually the key limiting factor though. The brakes are a huge 15".

What is "best" for the GT-R is clearly NOT best for the M3, why is that so hard to understand?

I don't understand how the 20's can be fine considering you run a 305 rear and a 255 front which is way off from the balance of the stock 245/265 . Is the car more prone to sudden oversteer with the 20's due to the front / rear size difference.?

Actually I run a 285 rear and 245 front. Oh, I see you've confused me with another poster. Nevermind, carry on...

A couple people on here have mentioned density. Can someone explain to me why density matters and not just mass? It doesn't matter how dense something is. For this example it only matters what it's mass is. Isn't it like that old kids game....Which is heavier a 1lb of bricks or a 1lb of feathers? At the end of the day it's the mass that affects the rotation right? Sorry if that is a stupid question. I'm obviously not an engineer.

Taking the "stick" example someone else mentioned I have another question. If 2 sticks are 13.365 inches long and 1 stick has an extra 2 lbs attached to the last 3 inches wouldn't that stick have greater "I", or inertia and be more difficult to rotate and/or stop rotation? Please don't answer with sidewall stiffness, lap times, predictable behavior of at the limit adhesion, etc. I understand all that, don't necessarily disagree, but quite frankly don't care.

This is a street setup on a car that is not my daily driver. I drive my Escalade during the week and race a 125cc Tony Kart on weekends. Other than practicing my heel-toe downshifting I don't race my M3 on the streets. My stock wheels/tires will be used for track duty. I'm merely just trying to satisfy a curiousity. Thanks.

A couple people on here have mentioned density. Can someone explain to me why density matters and not just mass? It doesn't matter how dense something is. For this example it only matters what it's mass is. Isn't it like that old kids game....Which is heavier a 1lb of bricks or a 1lb of feathers? At the end of the day it's the mass that affects the rotation right? Sorry if that is a stupid question. I'm obviously not an engineer.

Taking the "stick" example someone else mentioned I have another question. If 2 sticks are 13.365 inches long and 1 stick has an extra 2 lbs attached to the last 3 inches wouldn't that stick have greater "I", or inertia and be more difficult to rotate and/or stop rotation? Please don't answer with sidewall stiffness, lap times, predictable behavior of at the limit adhesion, etc. I understand all that, don't necessarily disagree, but quite frankly don't care.

This is a street setup on a car that is not my daily driver. I drive my Escalade during the week and race a 125cc Tony Kart on weekends. Other than practicing my heel-toe downshifting I don't race my M3 on the streets. My stock wheels/tires will be used for track duty. I'm merely just trying to satisfy a curiousity. Thanks.

Sure, I'll help. Mass and distance from the center of rotation are really the only things that matter, for both acceleration and angular acceleration. However, if in a given structure you replaced an equivalent volume of material A with a more dense volume of material B you would be altering both the mass and moment of inertia. It also helps when trying to do a mental estimation of the relative I between two structures, if you imagine sort of morphing one structure into another you can imagine the transformation being possible by replacing a less dense portion of it (say a section of tire) with a more dense (say aluminum) you can guesstimate the change in I. Again mass affects the resistance to linear (or rectilinear) acceleration just as moment of inertia resists spin up (angular acceleration).

You are right on with the stick example. In your example though one had a greater m and greater I. Here is how to take the understanding one step further. If you placed the extra 2 lb right at the center of the stick, call this condition A, and the extra 2 lb at the end of another identical stick, call this condition B. Both A and B would accelerate equally if you placed the same force on them (in such a way that you did not cause them to spin!). However, one will take more more torque to obtain a given angular acceleration.